Let $\Omega\subset \mathbb{R}^n$ an open set and $u:\Omega\to \mathbb{R}$ be a (locally) $L^1$-function. Then it is well known that the Lebesgue differentiation theorem holds: For almost every $x\in \Omega$, $$\frac{1}{|B_r(x)|} \int_{B_r(x)} (u(y) - u(x)) d y \to 0$$ as $r\to 0$. My question is if it is true that $$\frac{1}{|B_r(x)|} \int_{B_r(x)} |u(y) - u(x)| d y \to 0$$ as $r\to 0$ for almost every $x\in \Omega$. I am doubtful about this, but the universal counterexample in measure theory - the (characteristic function of the) fat Cantor set - doesn't work here. If it is actually true, then it could probably further be generalized to $$\frac{1}{|B_r(x)|} \int_{B_r(x)} |u(y) - u(x)|^p d y \to 0$$ if $u\in L^p(\Omega)$ and $p\ge 1$. Actually that is what I'm interested in.

It's true that $$\frac{1}{|B_r(x)|}\int_{B_r(x)} |u(y)-u(x)|^p\to 0$$ for $r\to 0$. For $p=1$ this is standard (Somehow I always overlooked the absolute value signs in the definition of Lebesgue points), and the same proof carries over for $p>1$ (at least the proof used in Ziemer, "Weakly differentiable functions", where variants of this theorem for Sobolev spaces are discussed as well).

Measure and integral. $\endgroup$ – Pietro Majer Oct 19 '11 at 17:05